A particle moves in the $xy$ -plane so that at any time $t\geq 0$ its coordinates are $x=2t^3+2$ and $y=-4t^4+4$. What is the particle's acceleration vector at $t=1$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $(6,48)$ (Choice B) B $(12,-48)$ (Choice C) C $(4,-24)$ (Choice D) D $(0,96)$
Solution: Background When solving motion problems, it's important to remember the relationship between the position vector $(x,y)$, the velocity vector $\vec{v}(t)$, and the acceleration vector $\vec{a}(t)$ of the moving particle: $\vec{v}(t)=\left(\dfrac{dx}{dt},\dfrac{dy}{dt}\right)$ $\vec{a}(t)=\dfrac{d}{dt}\vec{v}(t)=\left(\dfrac{d^2x}{dt^2},\dfrac{d^2y}{dt^2}\right)$ Setting up the math We are given that the particle's coordinates are $x=2t^3+2$ and $y=-4t^4+4$, which means its position vector is $(2t^3+2, -4t^4+4)$. We are asked to find the particle's acceleration vector at $t=1$. In other words, we need to find $\vec{a}(1)$. Finding $\vec{v}(t)$ $\begin{aligned} \vec{v}(t)&=\left(\dfrac{d}{dt}(2t^3+2),\dfrac{d}{dt}(-4t^4+4)\right) \\\\ &=\left(6t^2,-16t^3\right) \end{aligned}$ Finding $\vec{a}(t)$ $\begin{aligned} \vec{a}(t)&=\dfrac{d}{dt}\vec{v}(t) \\\\ &=\left(\dfrac{d}{dt}(6t^2),\dfrac{d}{dt}\left(-16t^3\right) \right) \\\\ &=\left(12t,-48t^2\right) \end{aligned}$ Finding $\vec{a}(1)$ $\begin{aligned} \vec{a}({1})&=\left(12(1),-48(1)^2\right) \\\\ &=(12,-48) \end{aligned}$ In conclusion, the particle's acceleration vector at $t=1$ is $(12,-48)$.